Here are the step-by-step solutions for the three problems on your worksheet.
Problem 1: The Shed Roof Beam
Step 1: Identify the shape.
The diagram shows a right-angled triangle formed by the roof beam, the width of the shed, and the height of the roof.
* The
width (base) is $5\text{ m}$.
* The
height is $1.3\text{ m}$.
* The
roof beam is the hypotenuse (the longest side).
Step 2: Use Pythagoras' Theorem.
The formula is $a^2 + b^2 = c^2$, where $c$ is the hypotenuse.
$$5^2 + 1.3^2 = c^2$$
Step 3: Calculate the squares.
$$5 \times 5 = 25$$
$$1.3 \times 1.3 = 1.69$$
Step 4: Add them together.
$$25 + 1.69 = 26.69$$
So, $c^2 = 26.69$.
Step 5: Find the square root.
To find the length of the beam ($c$), take the square root of $26.69$.
$$\sqrt{26.69} \approx 5.1662\dots$$
Step 6: Round to 1 decimal place (1dp).
The second decimal digit is $6$, which is $5$ or greater, so we round up the first digit.
$$5.166\dots \rightarrow 5.2$$
***
Problem 2: Captain's Geo-Cache Hunt
Step 1: Visualize the triangle.
Captain starts at a point, walks West, and then needs to go South to reach the treasure. This creates a right-angled triangle.
* The distance he walked West is one leg:
$24\text{ m}$.
* The straight-line distance from his starting point to the treasure is the hypotenuse:
$40\text{ m}$.
* The distance he needs to walk South is the missing leg (let's call it $b$).
Step 2: Set up Pythagoras' Theorem.
$$a^2 + b^2 = c^2$$
$$24^2 + b^2 = 40^2$$
Step 3: Calculate the squares.
$$24 \times 24 = 576$$
$$40 \times 40 = 1600$$
Step 4: Solve for $b^2$.
$$576 + b^2 = 1600$$
Subtract $576$ from $1600$:
$$b^2 = 1600 - 576$$
$$b^2 = 1024$$
Step 5: Find the square root.
$$\sqrt{1024} = 32$$
***
Problem 3: The Tower Anchor Line
Step 1: Identify the sides.
This is another right-angled triangle involving the tower, the ground, and the anchor line.
* The
tower height is a leg:
$96\text{ ft}$.
* The
anchor line is the hypotenuse:
$105\text{ ft}$.
* The
distance from the tower along the ground is the missing leg (let's call it $a$).
Step 2: Set up Pythagoras' Theorem.
$$a^2 + 96^2 = 105^2$$
Step 3: Calculate the squares.
$$96 \times 96 = 9216$$
$$105 \times 105 = 11025$$
Step 4: Solve for $a^2$.
$$a^2 + 9216 = 11025$$
Subtract $9216$ from $11025$:
$$a^2 = 11025 - 9216$$
$$a^2 = 1809$$
Step 5: Find the square root.
$$\sqrt{1809} \approx 42.532\dots$$
Step 6: Round to the nearest foot.
The decimal part is $.53$, which is less than $.5$? No, wait. $.53$ is greater than $.5$. Let's look closer.
Actually, standard rounding rules: if the decimal is $.5$ or higher, round up. If it is lower than $.5$, round down.
$42.532...$
The digit after the decimal is $5$. So we round up to $43$.
*Self-Correction Check:* Let's double check $\sqrt{1809}$.
$42^2 = 1764$
$43^2 = 1849$
$1809$ is closer to $1849$ (difference of $40$) than to $1764$ (difference of $45$).
Therefore, $43$ is the correct nearest whole number.
Final Answer:
1) 5.2m
2) 32m
3) 43ft
Parent Tip: Review the logic above to help your child master the concept of pythagorean theorem word problems worksheet answer key.
